Set them equal and take the log of both sides. Then you probably can pick out a few by inspection. This is supposed to be a precalc problem and they don't know much. If this is on the GRE, I wouldn't be surprised if the people who cooked up the problem made a mistake.
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toypajmeDec 21 '12 at 19:55

@toypajme, that was my first attempt. It didn't seem like it could lead anywhere. So I tried something else
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HawkDec 21 '12 at 19:58

Hint: Look at the plot of the curves on top of each other regarding your first suspicion.
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AmzotiDec 21 '12 at 20:02

3

The argument for $x\gt 0$ was not right. In the long run $e^x\gt x^{12}$. But $e^2\lt 2^{12}$. So there is a root between $0$ and $2$, and a root somewhere beyond $2$.
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André NicolasDec 21 '12 at 20:02

4 Answers
4

$x^{12} = 2^x$ (for $x$ real) is equivalent to: either $x = 2^{x/12}$ or $-x = 2^{x/12}$. Since $2^{x/12}$ is convex, its graph intersects a straight line in $0$, $1$ or $2$ points.

$-x$ is decreasing while $2^{x/12}$ is increasing, and $-x > 1 > 2^{x/12}$ for $x < -1$ while $-x < 2^{x/12}$ for $x > 0$, so there is exactly one real solution of $-x = 2^{x/12}$ and it is in the interval $-1 \le x \le 0$.

$x < 0 < 2^{x/12}$ for $x < 0$, $x < 2^{x/12}$ for sufficiently large $x$, while $x > 2^{x/12}$ at $x=2$, so there are two real solutions of $x = 2^{x/12}$, one with $0 < x < 2$ and one in $2 < x < \infty$.

I thought of it this way; near the origin, $2^x$ is relatively flat, while $x^{12}$ points up sharply. This gives two points of intersection. But, eventually, exponentials outgrow any polynomial, so there must be another point of intersection where $2^x$ outgrows $x^{12}$.

Actually, I didn't think of the third root at the time either and put down 2 as my answer on the practice exam. But hey, only 12 percent of test takers get it right!

This is how I did it. There seem to be a few problems like this on each GRE, where if you try do things rigorously it's impossible, but intuitively it's obvious once you see the point of the problem.
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PotatoSep 25 '14 at 1:04

First thing to observe is that both the functions are continuous and the graphs are "smooth" curves.

There is one and only one point of intersection in $(- \infty, 0)$ as $x^{12}$ is increasing & $2^x$ is decreasing as $x \rightarrow- \infty$ and $2^0 > 0^{12}$.

Now observe that when $x \rightarrow \infty, $ both the functions increase and $2^x $ dominates $x^{12}$ eventually. Since $ 2^{12} > 2^2 $, it must be the case that the graphs intersect at some point in $(2, \infty)$. This is the only point of intersection in that interval.

Also, $2^0 > 0^{12}$ and $2^{12} > 2^2 $ tells us that there is a point of intersection in $(0,2)$ and there is exactly one such intersection.